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cmizel · Apr 30, 2012, 05:27 PM

integrate e^(-(x^2))dx

jcaron2 · May 4, 2012, 07:02 AM

Originally Posted by cmizel

integrate e^(-(x^2))dx

There's no closed-form solution to the integral, though it can be written as an asymptotic series:

First, rewrite the integral into a form where you can do integration by parts:

$\int e^{-x^2}\;dx=\int -$\frac{1}{2x}$ \; 2x e^{-x^2}\;dx$

Now we can use the integration by parts formulation

$\int u \; dv = uv-\int v \; du$ ,

where

$u=-$\frac{1}{2x}$$

and

$dv = 2x e^{-x^2}\;dx$

That gives you

$\int e^{-x^2}\;dx= -$\frac{1}{2x}$e^{-x^2}-\int $\frac{1}{2x^2}$e^{-x^2} \; dx$

Now you have a closed-form term, plus a new integral. If you employ integration by parts to solve the new integral, you'll get yet another closed-form term and another new integral (this time with higher powers of x). If you keep going, integrating by parts over and over and over, you build up a series of closed-form terms, and end up with an integral whose value is approaching zero. In the limit, as the number of iterations goes to infinity, you end up with:

$\int e^{-x^2}\;dx=\frac{e^{-x^2}}{2}$x^{-1}-\frac12 x^{-3} + \frac34 x^{-5} - \frac{15}{8} x^{-7}+\frac{105}{16}x^{-9}- \; ... \;$ = \frac{e^{-x^2}}{2}\sum_{n=0}^\infty \frac{$-1$^n$2n-1$!!}{2^n}x^{2n+1}$